Problem: Nora uploaded a funny video on her website, which rapidly gains views over time. The following function gives the number of views $t$ days after Nora uploaded the video: V ( t ) = 100 ⋅ e 0.4 t V(t)=100\cdot e\^{ 0.4t} What is the instantaneous rate of change of the number of views $4$ days after the video was uploaded? Choose 1 answer: Choose 1 answer: (Choice A) A $198$ views (Choice B) B $198$ views per day (Choice C) C $495$ views (Choice D) D $495$ views per day
Explanation: Understanding the problem The instantaneous rate of change of $V(t)$ is given by its derivative, $V'(t)$. Therefore, the instantaneous rate of change of the number of views $4$ days after the video was uploaded is $V'(4)$. Let's find $V'(t)$ and evaluate it at $t=4$. Finding $V'(t)$ V ′ ( t ) = 40 ⋅ e 0.4 t V'(t)=40\cdot e\^{ 0.4t} Finding $V'(4)$ V ′ ( 4 ) = 40 ⋅ e 0.4 ( 4 ) = 40 ⋅ e 1.6 ≈ 198 \begin{aligned} V'(4)&=40\cdot e\^{ 0.4(4)} \\\\ &=40\cdot e\^{ 1.6} \\\\ &\approx 198 \end{aligned} Interpreting units $V(t)$ is the number of ${\text{views}}$, $t$ ${\text{days}}$ after the video was uploaded. Therefore, we measure its rate of change in ${\text{views}}$ per ${\text{day}}$. In conclusion, the instantaneous rate of change of the number of views, $4$ days after the video was uploaded, is $198$ views per day. The rate of change is positive because the number of views is increasing.